What is the theoretical yield of aluminum oxide if 2.20 of aluminum metal is exposed to 1.95 of oxygen?

(Assuming that the numbers given are moles)

For a problem like this, the best thing to do is write a balanced equation. This gets us:

4Al + 3O2 -> 2Al2O3

Now we must find the limiting reactant. That is, the reactant that will result in less product. Using the Coefficients as fractions, we find that 2.2mol of Al should yield:

(2.2/4)*2= 1.1mol Al2O3

Next we do the same with 1.95 mol of O2 to get:

(1.95/3)*2= 1.3mol Al2O3

We go with the lesser amount, so the answer is 1.1mol Al2O3.

Or 112g